Splitting field of $ f(x) = x^4 - 2 $ over $ \Bbb{Q} $ with minimal Galois theory and avoiding the use of embedding in $ \Bbb{C} $.

Question

Dummit and Foote Problem 13.4.1 (in Section 13.4 which simply proves the existence and very basic properties of splitting fields and algebraic closures) asks me to

Determine the splitting field and its degree over $ \Bbb{Q} $ for $ f(x) = x^4 - 2 $.

Additional Context and my actual question: I'm well aware that the splitting field is $ \Bbb{Q}(\sqrt[4]{2},i) $ with degree $ 8 $ over $ \Bbb{Q} $ but would like to prove it "properly" and from the basic principles of field extension theory discussed so far in the text. I am loath [edit, not loathe] to use any reference to embedding in $ \Bbb{R} $ or $ \Bbb{C} $ since the fundamental theorem of algebra is unproven by this point in the text and at any rate seems too powerful and not in the spirit of this question or section. I propose a solution below. Please read it and tell me if I missed something far simpler.

What I've done so far: $ f(x) $ is Eisenstein at $ 2 $ so is irreducible. Let $ \alpha $ be a root. In the spirit of doing things "the right way" I'd like to insist on not calling it $ \sqrt[4]{2} $ so as to avoid smuggling in any notion of embedding into $\Bbb{R}$ or $\Bbb{C}$. Now $ \Bbb{Q}(\alpha) $ is a degree 4 extension over $ \Bbb{Q} $ containing two roots $ \alpha $ and $ -\alpha $ of $ f(x) $. I need to ascertain that this extension does not contain any more roots. I can factor $ f(x) = (x-\alpha)(x+\alpha)(x^2 + \alpha^2) $, so my task would be to check that $ x^2 + \alpha^2 $ is irreducible in $ \Bbb{Q}(\alpha) $. In fact if I can show $ x^2 + 1 $ is irreducible then this result would follow since, if $ \beta $ is a root of $ x^2 + 1 $ then $ \alpha \beta $ would be a root of $ x^2 + \alpha^2 $. But it's not immediately clear why $ x^2 + 1 $ should be irreducible in $ \Bbb{Q}(\alpha) $ (again, I emphasize, using purely algebraic notions and avoiding imbedding into $ \Bbb{C} $).

After several attempts, here's the simplest argument I could come up with: consider the degree 2 extension $ E = \Bbb{Q}(\alpha^2) $. In $ E $, $ f(x) $ factors $$ f(x) = (x^2 - \alpha^2)(x^2 + \alpha^2) $$ and we know these factors are irreducible (or else it would contradict that $ \alpha $ is degree $4$). Letting $ K $ be the splitting field of $ f(x) $ over $E$, it suffices to show that $ K $ is a biquadratic extension of $ E $. I've solved this problem

Problem 13.2.8: Let $ F $ be a field of characteristic $ \neq 2 $. Let $ D_1, D_2 \in F $ where none of $ D_1 $, $ D_2 $, and $ D_1 D_2 $ are perfect squares. Then $ F(\sqrt{D_1}, \sqrt{D_2}) $ is a degree 4 extension, called a biquadratic extension.

I can apply this with $ F = E = \Bbb{Q}(\alpha^2) $, $ D_1 = \alpha^2 $, $ D_2 = -\alpha^2 $. Indeed $ D_1 D_2 = -2 $ is readily checked (by direct computation) to not be a perfect square in $ E = \{ a + b \alpha^2 \mid a,b \in \Bbb{Q} \} $. I conclude that $$ K = E(\alpha, i\alpha) = \Bbb{Q}(\alpha^2)(\sqrt{D_1}, \sqrt{D_2}) $$ is a biquadratic extension of $ E = \Bbb{Q}(\alpha^2) $ and conclude that the splitting field of $ x^4 - 2 $ is a degree $ 8 $ extension of $ \Bbb{Q} $ generated by $ \alpha $ (a root of $ x^4 - 2 $) and $ \beta $ (a root of $ x^2 + 1 $).

A meta-question, if you'll indulge me: Proving Fundamental Theorem of Algebra is not hard to do with a bit of topology. I could easily write down a proof using winding number. For a student like me, what should be my philosophy toward using embedding into $ \Bbb{C} $ to think about and solve problems in field extension theory? On the one hand, using overly-powerful machinery may cause me to miss more general techniques and subtleties of Galois theory (such as over finite fields). On the other hand, $ \Bbb{C} $ supplies such rich intuition. A great example of this is that the embedding into $ \Bbb{C} $ makes it "obvious" that there exists a primitive $ n^{\text{th}} $ root of unity $ \zeta_n $ which we can identify with $ \exp(i\tau /n) $, whereas proving this purely algebraically is not-at-all straightforward. It also would have made this problem so much easier. Was this even a useful exercise, working so hard to avoid $ \Bbb{C} $, or was my intution (that I should even look for such a solution) misguided?

Answer 1

The part of the argument that uses embeddings into $\Bbb R$ and $\Bbb C$ here that is essential is that a field that does not admit an ordering cannot embed into a field that does admit an ordering, because a finite extension of $\Bbb Q$ embeds into $\Bbb R$ iff it admits an ordering. This adresses the meta-question as well, because we can always argue in terms of orderings instead of arguing with embeddings into $\Bbb R$. This is a slight digression from the usual treatments of field theory, but it's an interesting piece of mathematics on its own.

The general theory of ordered fields is purely algebraic, it does not rely on Galois theory and we don't need to mention $\Bbb R$ or $\Bbb C$.

Let us introduce some definition first:

Definition A field $K$ is called formally real if there exists an order on it making it an ordered field.

Note that instead of specifying the ordering as a relation, one can also specify the (so called) cone of positive elements. A positive cone $P$ on field $K$ is a set $P$ satisfying

• $P+P \subseteq P$
• $P\cdot P \subseteq P$
• $P \cap -P = \{0\}$
• $P \cup -P=K$

Often it is useful to start with a smaller set and then extend it to positive cone. A straightforward Zorn's lemma argument shows the following

Lemma Let $P$ be a subset of a field $K$ such that $P+P \subseteq P$, $P \cdot P \subseteq P$, $-1 \notin P$ and $K^2 \subseteq P$, where $K^2$ denotes the set of squares in $K$. Then $P$ can be extended to a positive cone.

We want to prove the following lemma to show without reference to $\Bbb R$ that $\Bbb Q(\alpha)$ is formally real.

Lemma Let $K$ be a formally real field and let $a \in K$ such that $a$ is positive in $K$ for some ordering $\leq$. Then $\leq$ extends to an ordering of $K(\sqrt{a})$, which is thus formally real.

Proof Let $P = \{k \in K : k\geq 0\}$ and $Q=\left\{\sum_{i=1}^n p_iv_i^2 : n \in \Bbb N, p_i \in P, v_i \in K(\sqrt{a}) \right\}$. Then clearly $K(\sqrt{a})^2 \subset Q$, $Q+Q \subset Q$, $Q \cdot Q \subset Q$. So the hard part is showing $-1 \notin Q$. Suppose that $-1=\sum_{i=1}^n p_i (x_i+y_i\sqrt{a})^2$ for $p_i \in P, x_i,y_i \in K$. Multiplying this out and comparing coefficients, we get $-1=\sum_{i=1}^np_i(x_i^2+ay_i^2)\in P$. This is a contradiction and we win. Now we can apply the first lemma.

With this tool at hand, we can show that $\Bbb Q(\alpha)$ is formally real. First, $\Bbb Q(\sqrt{2})/\Bbb Q$ is formally real, because $\Bbb Q$ can be ordered and $2$ is positive. Choose an ordering on $\Bbb Q(\sqrt{2})$. Now in $\Bbb Q(\sqrt{2})$ one of $\sqrt{2}$ and $-\sqrt{2}$ must be positive for the chosen ordering. Let $\beta$ be the positive root of $X^2-2$. Then by the lemma $\Bbb Q(\sqrt{\beta})$ is formally real. But $\sqrt{\beta}$ is a root of $X^4-2$, so this is the field that we want. To finish it all off, when we have a formally real field, then $-1$ is not a sum of squares, let alone a square, so $X^2+1$ has no roots and is irreducible over any formally real field.

Answer 2

Here's my philosophy: to solve this problem, one shouldn't use facts that haven't been proven yet in the text. You especially shouldn't use any facts from other fields of math --- you're just hindering your algebraic development. However, a student should be able to use any intuition they've so far acquired in their learning path. So, here's a solution I believe the average student at this level has the understanding to come up with.

In fact if I can show $x^2+1$ is irreducible then this result would follow

Starting from here, I believe it's reasonably intuitive to attempt to develop the notion of a conjugate in $\mathbb{Q}(\alpha)$. So, define $$\overline{a + b\alpha + c\alpha^2 + d\alpha^3} := a - b\alpha + c\alpha^2 - d\alpha^3.$$ I've simply sent $\alpha$ to $-\alpha$, so it's completely analogous to the familiar concept of complex conjugation. Then it can be checked that $z^2 + 1 =: f(z) = 0$ iff $f(\bar{z}) = 0$ by purely computational means. Then, assuming $f$ is reducible in $\mathbb{Q}(\alpha)$, $f(x) = (x - a - b\alpha - c\alpha^2 - d\alpha^3)(x - a + b\alpha - c\alpha^2 + d\alpha^3)$. Again, by pure computation

$$x^2 + 1 = f(x) = x^2 + (-2a - 2c\alpha^2)x + a^2 + 2c^2 - 4bd + (2ac - b^2 - 2d^2)\alpha^2$$

and so $c\alpha^2 = a$, from which $a = c = 0$ since $\alpha$ is of degree $4$. Then $b^2 + 2d^2 = 0$ from which $b = d = 0$ since $b, d$ are rational. But that's clearly not a root.

The ideas here are related to Galois theory, but I think that's the point of the exercise, to get you to develop some introductory approaches by yourself. Properties of $\mathbb{C}$ aren't actually utilized, they just serve as motivation. We could've also applied the same technique to the polynomial $x^2 + \alpha^2$, so your simplification wasn't strictly necessary, although it's perhaps easier to spot the method of finding the conjugate when the polynomial has purely rational coefficients.

Answer 3

Edit: the below answer is wrong. Everything is correct except the claim that $ (2) $ is a prime ideal in $ \Bbb{Z}[i] $. After all, $ (1+i)(1-i) = 2 $. Thus I cannot use the Eisenstein criterion to claim that $ x^4 - 2 $ is irreducible over $ \Bbb{Z}[i] $. I hope my footnote (checking that all pairwise products of the linear factors $ (x-\alpha)(x+\alpha)(x-i \alpha)(x+i \alpha) $ don't belong to $ \Bbb{Q}(i) $) suffices to salvage my proof.

Original Answer: I found a somewhat satisfying answer to my own question that was different enough from the accepted answer I thought it was worth posting. I prefer it as a direct proof with no use of contradiction or muddling computation.

Let $ f(x) = x^4 - 2 \in \Bbb{Q}[x] $. Fix a splitting field $ K $ over $ \Bbb{Q} $. Let $ \alpha \in K $ be a root of $ f(x) $. We have $ f(x) = (x - \alpha)(x + \alpha)(x^2 + \alpha^2) $. Since this splits over $ K $, pick a root $ \alpha' \in K $ of $ x^2 + \alpha^2 $. Then $(\alpha/\alpha')^2 = -1 $. Let $ \beta = \alpha/\alpha' $. We see that $ \Bbb{Q}(\alpha, \beta) = K $ since $ f(x) $ splits $$ f(x) = (x - \alpha)(x + \alpha)(x - \beta\alpha)(x + \beta\alpha) $$ Our goal is to show that $ [K : \Bbb{Q}] = 8 $. It can only be 4 or 8 depending on whether $ \beta \in \Bbb{Q}(\alpha) $. But $ f(x) $, viewed as an element of $ \Bbb{Z}[i][x] $ is irreducible since it is Eisenstein at the prime ideal $ (2) \subset \Bbb{Z}[i] $, and hence is irreducible in $ \Bbb{Q}(i) \cong \Bbb{Q}(\beta) $ by Gauss' Lemma (Section 9.3, Proposition 5). So to conclude, $$ [K:\Bbb{Q}] = [\Bbb{Q}(\beta, \alpha) : \Bbb{Q}(\beta)][\Bbb{Q}(\beta):\Bbb{Q}] = 4 \cdot 2 = 8 $$ I don't usually write this, but Q.E.D.*

*And if one is uneasy about the use of $ \Bbb{Z}[i] $ (what is Gauss' Lemma, how do you know $ \Bbb{Q}(i) $ is the quotient ring, why is $ (2) $ prime, etc.), one can instead systematically verify that none of the factors $ (x-\alpha) $, $ (x + \alpha) $, $ (x - \beta\alpha) $, $ (x + \beta\alpha) $, nor any of their pairwise products, belong to $ \Bbb{Q}(\beta)[x] $. This suffices to guarantee that $ f(x) $ is irreducible over $ \Bbb{Q}(\beta) $.